Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 1;
const int MOD = 1e9 + 7;
const int dx[] = {+1, -1, 0, 0};
const int dy[] = {0, 0, +1, -1};
const int N = 1e5 + 5;
void frepik() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
int n, m, q, up[N][20], tin[N], tout[N], timer;
vector<int> g[N], d[N][20], a[N];
vector<int> merge(vector<int> l, vector<int> r) {
  vector<int> res;
  int p1 = 0, p2 = 0;
  for (int i = 0; i < min(10, (int)l.size() + (int)r.size()); i++) {
    if (p1 == l.size())
      res.push_back(r[p2++]);
    else if (p2 == r.size())
      res.push_back(l[p1++]);
    else {
      if (l[p1] < r[p2])
        res.push_back(l[p1++]);
      else
        res.push_back(r[p2++]);
    }
  }
  return res;
}
void dfs(int first, int pr) {
  tin[first] = ++timer;
  for (int to : g[first]) {
    if (to != pr) {
      up[to][0] = first;
      d[to][0] = a[first];
      dfs(to, first);
    }
  }
  tout[first] = ++timer;
}
bool upper(int u, int v) { return (tin[u] <= tin[v] && tout[u] >= tout[v]); }
vector<int> lca(int u, int v) {
  if (u == v) return a[v];
  vector<int> res = merge(a[v], a[u]);
  for (int i = 19; i >= 0; i--) {
    if (up[v][i] != -1 && !upper(up[v][i], u)) {
      res = merge(res, d[v][i]);
      v = up[v][i];
    }
  }
  for (int i = 19; i >= 0; i--) {
    if (up[u][i] != -1 && !upper(up[u][i], v)) {
      res = merge(res, d[u][i]);
      u = up[u][i];
    }
  }
  if (!upper(u, v) && !upper(v, u)) res = merge(res, a[up[v][0]]);
  return res;
}
int main() {
  frepik();
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int first, second;
    cin >> first >> second;
    g[first].push_back(second);
    g[second].push_back(first);
  }
  for (int i = 1; i <= m; i++) {
    int v;
    cin >> v;
    a[v].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(a[i].begin(), a[i].end());
    while (a[i].size() > 10) a[i].pop_back();
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 20; j++) up[i][j] = -1;
  dfs(1, -1);
  for (int j = 1; j <= 20; j++) {
    for (int i = 1; i <= n; i++) {
      if (up[i][j - 1] != -1 && up[up[i][j - 1]][j - 1] != -1) {
        up[i][j] = up[up[i][j - 1]][j - 1];
        d[i][j] = merge(d[up[i][j - 1]][j - 1], d[i][j - 1]);
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    int v, u, k;
    cin >> v >> u >> k;
    vector<int> res = lca(v, u);
    k = min(k, (int)res.size());
    cout << k << " ";
    for (int l = 0; l < k; l++) cout << res[l] << " ";
    cout << endl;
  }
  return 0;
}